To solve the problem step by step, we will follow the principles of buoyancy and the relationship between density, volume, and weight.
### Step 1: Calculate the Volume of the Hollow Sphere
The volume \( V \) of the hollow sphere can be calculated using the formula for the volume of a sphere:
\[
V = \frac{4}{3} \pi (R^3 - r^3)
\]
where \( R \) is the outer radius and \( r \) is the inner radius. Given:
- Outer radius \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Inner radius \( r = 9 \, \text{cm} = 0.09 \, \text{m} \)
Calculating the volume:
\[
V = \frac{4}{3} \pi (0.1^3 - 0.09^3)
\]
\[
= \frac{4}{3} \pi (0.001 - 0.000729)
\]
\[
= \frac{4}{3} \pi (0.000271) \approx 0.000454 \, \text{m}^3
\]
### Step 2: Calculate the Weight of the Hollow Sphere
The weight \( W \) of the hollow sphere can be expressed as:
\[
W = V \cdot \rho_m \cdot g
\]
where \( \rho_m \) is the density of the material of the sphere and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
### Step 3: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the sphere when it is half-submerged can be calculated using the volume of the liquid displaced, which is half the volume of the sphere:
\[
F_b = \frac{1}{2} V_l \cdot \rho_l \cdot g
\]
where \( V_l \) is the volume of the liquid displaced and \( \rho_l \) is the density of the liquid. Given the specific gravity of the liquid is \( 0.8 \):
\[
\rho_l = 0.8 \cdot 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3
\]
Thus, the buoyant force can be expressed as:
\[
F_b = \frac{1}{2} \cdot \frac{4}{3} \pi (0.1^3) \cdot 800 \cdot g
\]
### Step 4: Set Up the Equation for Equilibrium
At equilibrium, the weight of the sphere is equal to the buoyant force:
\[
W = F_b
\]
Substituting the expressions for \( W \) and \( F_b \):
\[
\frac{4}{3} \pi (0.1^3 - 0.09^3) \cdot \rho_m \cdot g = \frac{1}{2} \cdot \frac{4}{3} \pi (0.1^3) \cdot 800 \cdot g
\]
### Step 5: Simplify and Solve for Density
Canceling common terms:
\[
(0.1^3 - 0.09^3) \cdot \rho_m = \frac{1}{2} \cdot (0.1^3) \cdot 800
\]
Calculating \( 0.1^3 - 0.09^3 = 0.001 - 0.000729 = 0.000271 \):
\[
0.000271 \cdot \rho_m = \frac{1}{2} \cdot 0.001 \cdot 800
\]
\[
0.000271 \cdot \rho_m = 0.4
\]
\[
\rho_m = \frac{0.4}{0.000271} \approx 1474.5 \, \text{kg/m}^3
\]
### Step 6: Calculate the Density of a Liquid for Complete Submersion
For the sphere to be completely submerged, the buoyant force must equal the weight of the sphere:
\[
F_b = W
\]
Using the entire volume of the sphere:
\[
\frac{4}{3} \pi (0.1^3) \cdot \rho_l \cdot g = \frac{4}{3} \pi (0.1^3 - 0.09^3) \cdot \rho_m \cdot g
\]
Cancelling out common terms and solving for \( \rho_l \):
\[
\rho_l = \frac{(0.1^3 - 0.09^3)}{(0.1^3)} \cdot \rho_m
\]
Substituting the values:
\[
\rho_l = \frac{0.000271}{0.001} \cdot 1474.5 \approx 400 \, \text{kg/m}^3
\]
### Final Answers
- The density of the material of the sphere is approximately \( 1474.5 \, \text{kg/m}^3 \).
- The density of the liquid in which the hollow sphere would just float completely submerged is \( 400 \, \text{kg/m}^3 \).
